How Do You Approach Reduction of Order for This Differential Equation?

[Matt Jacques]

Although on the whole my Boyce & DiPrima book is fairly good in of itself and compared with others, but of-coarse some areas are lacking.

vt'' - 3v' = 0

I suppose I integrate twice, but then what?

 

[matt grime]

vt''-3v'? You might want to check that, v and t are both functions of some paramater?

 

[Matt Jacques]

More specifically, v(t)*t'' - 3v(t) = 0

I'd go insane if I kept that (t) in there while differentiating and combining. :P

 

[matt grime]

And that factorizes as:

v(t)*(t''-3)=0

not sure that's right either as now the second v is not primed.

 

[Matt Jacques]

v''(t)*t - 3v'(t) = 0

Sorry, this is it now.

 

[matt grime]

Well, one solution is v(t)=t^3, don't you have to use this to get the other solution?


Edit, of course it isn't. There's the constant solution, duh, and that gives the answer in the next post.

 

[Max0526]

The general solution

Hi;
The general solution is:
v(t)=C_1t^4+C_2
Good luck,
Max.

 

[bowzerman]

add the 3v' to the other side getting:
vt''=3v'
divide v to the other side and then integrate with respect to t and v, based on the side of the equation, being sure to keep the integration constants in.

 

[HallsofIvy]

add the 3v' to the other side getting:
vt''=3v'
divide v to the other side and then integrate with respect to t and v, based on the side of the equation, being sure to keep the integration constants in.

No, you cannot treat second derivatives like that. If you really had a single equation in two dependent variables, you will need another equation to get an explicite solutions just as you cannot solve one numeric equation in two variables.

The problem is, after editing, tv"j- 3v'= 0.
Matt Jacques, since you explicitely refer to "reduction of order", I think you mean this: set u= v' so that u'= v" and you have the first order equation tu'- 3u= 0. Now, you can do exactly what bowzerman suggested, since this is now a first order equation.
t du/dt= 3u so du/u= 3t dt. ln|u|= (3/2)t²+ C and then
u= C_1 e^{(3/2)t^2.<br /> Since u= dv/dt, we have<br /> dv= C_1 e^{3/2}t^2 dt <br /> and <b>only</b> have to integerate again. I'll leave that to you!

 

[HallsofIvy]

add the 3v' to the other side getting:
vt''=3v'
divide v to the other side and then integrate with respect to t and v, based on the side of the equation, being sure to keep the integration constants in.

No, you cannot treat second derivatives like that. If you really had a single equation in two dependent variables, you will need another equation to get an explicite solutions just as you cannot solve one numeric equation in two variables.

The problem is, after editing, tv"- 3v'= 0.
Matt Jacques, since you explicitely refer to "reduction of order", I think you mean this: set u= v' so that u'= v" and you have the first order equation tu'- 3u= 0. Now, you can do exactly what bowzerman suggested, since this is now a first order equation.
t du/dt= 3u so du/u= 3t dt. ln|u|= (3/2)t²+ C and then
u= C_1 e^{(3/2)t^2.<br /> Since u= dv/dt, we have<br /> dv= C_1 e^{3/2}t^2 dt <br /> and <b>only</b> have to integerate again. I'll leave that to you!

 

[bowzerman]

yeah, sorry, I was just wrote that one too quick. Forgot to clarify.

 

[bowzerman]

oh, and shouldn't that be:
t du/dt = 3u
1/u du = 3/t ?
ln(u)=3ln(t) +C
u= c*t^3

, then sub, integrate, and then plug in?